complex infinite series

[logistic_guy]

here is the question

Use multiplication of series to show that \(\displaystyle \frac{e^z}{z(z^2 + 1)} = \frac{1}{z} + 1 - \frac{1}{2}z - \frac{5}{6}z^2 + \cdots \ (0 < |z| < 1)\).


my attemb
i think it want me to find infinite series for \(\displaystyle e^z\) and \(\displaystyle \frac{1}{z}\) and \(\displaystyle \frac{1}{z^2 + 1}\) then multiply them together
i know how to find taylor series for \(\displaystyle e^z\)
but i don't know how to find series for \(\displaystyle \frac{1}{z}\) and \(\displaystyle \frac{1}{z^2 + 1}\)

 

[fresh_42]

You can leave [imath] 1/z [/imath] as it is. For [imath] 1/(1+z^2) [/imath] use long division or look it up on Wolfram Alpha, or build the Taylor series by differentiations.

 

[logistic_guy]

You can leave [imath] 1/z [/imath] as it is. For [imath] 1/(1+z^2) [/imath] use long division or look it up on Wolfram Alpha, or build the Taylor series by differentiations.

i don't understand what you mean by long division
how to do division when i've only \(\displaystyle 1\) in the numator?
i thought long division is done when the numator have higher degree

 

[topsquark]

i don't understand what you mean by long division
how to do division when i've only \(\displaystyle 1\) in the numator?
i thought long division is done when the numator have higher degree

Google polynomial division.

-Dan

 

[fresh_42]

i don't understand what you mean by long division
how to do division when i've only \(\displaystyle 1\) in the numator?
i thought long division is done when the numator have higher degree

I don't care. Nike: Just do it!

[math]\begin{array}{lll} 1:(1+z^2)=1 - z^2 + z^4\mp \text{ etc.}\\[6pt] -(1+z^2)\\[6pt] ----------\\[6pt] -z^2\\[6pt] -(-z^2-z^4)\\[6pt] -------------\\[6pt] z^4\\[6pt] -(z^4+z^6)\\[6pt] -------------\\[6pt] -z^6\\[6pt] \text{etc.} \end{array}[/math]
The only thing that cannot be divided by is zero.

 

[logistic_guy]

Google polynomial division.

-Dan

i know how to do it but not with \(\displaystyle 1\) in the numator

I don't care. Nike: Just do it!

[math]\begin{array}{lll} 1:(1+z^2)=1 - z^2 + z^4\mp \text{ etc.}\\[6pt] -(1+z^2)\\[6pt] ----------\\[6pt] -z^2\\[6pt] -(-z^2-z^4)\\[6pt] -------------\\[6pt] z^4\\[6pt] -(z^4+z^6)\\[6pt] -------------\\[6pt] -z^6\\[6pt] \text{etc.} \end{array}[/math]
The only thing that cannot be divided by is zero.

why the first step in the long division, you divide \(\displaystyle 1\) by \(\displaystyle 1\) and not \(\displaystyle 1\) by \(\displaystyle z^2\)?

 

[topsquark]

i know how to do it but not with \(\displaystyle 1\) in the numator


why the first step in the long division, you divide \(\displaystyle 1\) by \(\displaystyle 1\) and not \(\displaystyle 1\) by \(\displaystyle z^2\)?

Honestly. You've never seen this before? Have you taken a course in complex Calculus?

-Dan

 

[fresh_42]

i know how to do it but not with \(\displaystyle 1\) in the numator


why the first step in the long division, you divide \(\displaystyle 1\) by \(\displaystyle 1\) and not \(\displaystyle 1\) by \(\displaystyle z^2\)?

Yes, because diving by [imath] z^2 [/imath] leads nowhere. The division by [imath] 1 [/imath] keeps it going. I can divide by whatever I want to, except zero, of course.

You can also formalize it by the Euclidean algorithm:
[math]\begin{array}{lll} 1&=1\cdot (1+z^2) - z^2\\[6pt] -z^2&=-z^2\cdot (1+z^2) +z^4\\[6pt] z^4&=z^4\cdot (1+z^2)-z^6\\[6pt] -z^6&=-z^6\cdot (1+z^2) +z^8\\[6pt] \text{etc.} \end{array}[/math]but the long division is easier and more known. The algorithm of dividing a polynomial [imath] p(z) [/imath] by a polynomial [imath] q(z) [/imath]
is always [imath] p(z)=s(z)\cdot q(z)+r(z) [/imath] no matter what [imath] p(z),q(z) [/imath] are. The example is a bit of a downside-up procedure in this specific case since the degree of [imath] r(z) [/imath] becomes larger rather than smaller. This means mathematically that the series is only a valid approximation for small values of [imath] z [/imath] around [imath] z=0. [/imath]

If you look at the thread you might observe that I already changed from your [imath] z^2+1 [/imath] in post #1 to my [imath] 1+z^2 [/imath] in post #2. That was on purpose!

In case you insist of the usual measure that the degree of the remainder [imath] r(z) [/imath] in Euclid's algorithm becomes smaller and smaller, then you have to transform the variable [imath] z \mapsto 1/x [/imath] and divide
[math] \dfrac{1}{1+z^2}=\dfrac{1}{1+\dfrac{1}{x^2}}=\dfrac{1}{\dfrac{x^2+1}{x^2}}=\dfrac{x^2}{x^2+1} [/math]instead and re-transform it by [imath] x\mapsto 1/z [/imath] afterward. I simply avoided these unnecessary steps by dividing [imath] 1:(1+z^2). [/imath]